What is the value of k if, #y=1/a^(1-log_a(x))# , #z=1/a^(1-log_a(y))# then #x=a^k# ?

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Answer 1 · 2017-10-22T00:38:36

# k = 1+log_a(y) #

We have:

# y = 1/a^(1-log_a(x)) # ..... [A]
# z = 1/a^(1-log_a(y)) # ..... [B]
# x = a^k \ \ \ \ \ \ \ \ \ \ \ \ \ \ # ..... [C]

From Eq [C] we have:

# x = a^k => k = log_ax #

From Eq [A] we have:

# a^(1-log_a(x)) = 1/y #
# :. 1-log_a(x) = log_a(1/y) #
# :. 1-log_a(x) = -log_a(y) #
# :. log_a(x) = 1+log_a(y) #

Thus:

# k = 1+log_a(y) #

Answer 2 · 2017-10-22T03:01:30

We have:

# y = 1/a^(1-log_a(x)) #

#=>y=a^(log_a(x)-1)#

#=>y=a^(log_a(x)-log_a a)#

#=>y=a^(log_a(x/a))#

#=>y=x/a#

#=>x=ay#

Inserting #x=a^k# (given) we get

#a^k=ay#

#=>log_a(a^k)=log_a(ay)#

#=>klog_a(a)=log_a(a)+log_a(y)#

#=>k=1+log_a(y)#